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Set forms a basis, and span help

  1. Feb 20, 2009 #1
    1. The problem statement, all variables and given/known data
    http://img16.imageshack.us/img16/6606/50381320.jpg [Broken]

    2. Relevant equations
    Please see above picture

    3. The attempt at a solution
    I believe for question a) I just need to add up all the matrices and then row reduce to RREF, which gives me:

    or Do I row reduce each matrices?

    I'm really not sure how to do b), any tips would be great!

    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 20, 2009 #2


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    think about linear combinations...

    in a) can every N[tex]\in[/tex] M23 be written as a linear combination of the matricies given?

    you could also think about what a simpler basis for M23 is and how they're related

    similarly in b) can the equation be written as a linear combination of the basis equations?
    Last edited: Feb 21, 2009
  4. Feb 21, 2009 #3
    I only wish to elaborate on Lanedance's post with a systematic spin.

    Think of the matrices in the set S as vectors in the vector space [tex]M_{23}[/tex] (the set of all 2x3 matrices). A basis for [tex]M_{23}[/tex] is
    [tex]\mathbf{e_1} = \begin{pmatrix}1 & 0 & 0\\ 0&0&0\end{pmatrix},[/tex] [tex]\mathbf{e_2} = \begin{pmatrix}0 & 1 & 0\\ 0&0&0\end{pmatrix}, \ldots, [/tex] [tex]\mathbf{e_6} = \begin{pmatrix}0 & 0 & 0\\ 0&0&1\end{pmatrix}.[/tex]

    The first vector in S can then be written as [tex]\mathbf{v_1} = [1 \; 2 \; 3 \; 4 \; 5 \; 6]^T[/tex] with respect to this basis, and similarly for the others, e.g., [tex]\mathbf{v_6} = [12 \; 10 \; 12 \; 12 \; 20 \; 18]^T[/tex].

    So to systematically determine whether the vectors in S are linearly independent (and hence form a basis), i.e., to determine whether the equation [tex]k_1\mathbf{v_1} + \ldots + k_6\mathbf{v_6}=\mathbf{0}[/tex] has a nontrivial solution, one can row reduce the corresponding matrix [tex][\mathbf{v_1} \ldots \mathbf{v_6}][/tex].

    It suffices to show that S spans the set of all such (i.e., of degree 4 or less) polynomials, which amounts to determining whether p1, p2, ..., p5 are linearly independent. Taking e1 = 1, e2 = x, ..., e5 = x^4, this is just like the part (a).

    Of course, in part (a) for example, if you can spot a non-trivial linear combination of the [tex]\mathbf{v_i}'s[/tex] that sums to [tex]\mathbf{0}[/tex], you are done. (I don't think it will be helpful if someone gives you one straight off the bat.)
  5. Feb 21, 2009 #4


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    if you do look at the large matrix suggested by unco, its also worth considering the determinant and how it relates to linear independence and row reduction
  6. Feb 21, 2009 #5
    http://img239.imageshack.us/img239/6250/picture2.png [Broken]
    Sorry, but I do not know how to use Tex. And thanks for the tips!

    From this, a) does not form the basis, and for b) p(x) belongs to the span (not too sure about b) as it has a row of zeros)
    Last edited by a moderator: May 4, 2017
  7. Feb 21, 2009 #6
    Right,though the last column of zeros is superfluous.

    This time taking vectors in row form (which is fine), the matrix you have here (albeit again with superfluous column of zeros) is what you would set up to see if {p, p1, p2, ..., p5} is linearly independent. You have shown they are not. But this does not necessarily mean p is a linear combination of p1, ... p5. For example, consider the set {(1,0), (2,0), (0,1)}: it is linearly dependent but (0,1) is not a linear combination of (1,0) and (2,0).

    Rather, as said earlier, if you verify that S={p1, p2, ..., p5} is linearly independent (just as you did in part (a)), then this set forms a basis for the set of polynomials, so span(S) = the whole set and so certainly p belongs to span(S).
  8. Feb 22, 2009 #7
    Thanks for the reply Unco!
    from matlab:
    http://img156.imageshack.us/img156/582/set.jpg [Broken]
    so P(x) belongs to the span of S. as it has a unique solution.
    Last edited by a moderator: May 4, 2017
  9. Feb 22, 2009 #8
    The first 5 columns of your row-reduced matrix show that S forms a basis. This is enough.

    The last (sixth) column does give you the coefficients of the linear combination of p1, ..., p5 giving p.

    The entire 5x6 matrix is the augmented matrix for the system

    [tex][\mathbf{p_1} \, \ldots \, \mathbf{p_5}][k_1 \, \ldots \, k_5]^T = \mathbf{p},[/tex]

    where [tex]k_i[/tex] is the coefficient of [tex]\mathbf{p_i}[/tex] in such a linear combination, and one wishes to solve for [tex][k_1 \, \ldots \, k_5]^T[/tex].

    The point is that not all the coefficients of the linear combination, i.e., the k_i, are non-zero.
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